What is the derivative of f(x)=(x^3-6x^2+3)/(x^2) ?

The derivative of f(x)=(x^3-6x^2+3)/(x^2) is (x^3-6)/(x^3)

What is the first derivative of f(x)=(x^3-6x^2+3)/(x^2) ?

The first derivative of f(x)=(x^3-6x^2+3)/(x^2) is (x^3-6)/(x^3)

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derivative of

f (x) = \frac{x ^{3} - 6 x ^{2} + 3}{x ^{2}}

\frac{x ^{3} - 6}{x ^{3}}

Solution steps

\frac{d}{d x} (\frac{x ^{3} - 6 x ^{2} + 3}{x ^{2}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( x ^{3} - 6 x ^{2} + 3 ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( x ^{3} - 6 x ^{2} + 3 )}{( x ^{2} ) ^{2}}

\frac{d}{d x} (x^{3} - 6 x^{2} + 3) = 3 x^{2} - 12 x

\frac{d}{d x} (x^{2}) = 2 x

= \frac{( 3 x ^{2} - 12 x ) x ^{2} - 2 x ( x ^{3} - 6 x ^{2} + 3 )}{( x ^{2} ) ^{2}}

Simplify

\frac{( 3 x ^{2} - 12 x ) x ^{2} - 2 x ( x ^{3} - 6 x ^{2} + 3 )}{( x ^{2} ) ^{2}} : \frac{x ^{3} - 6}{x ^{3}}

= \frac{x ^{3} - 6}{x ^{3}}

\int tan^{3} (θ) sec^{3} (θ) d θ

y^{^{'}} = - \frac{2 x}{( 1 + x ^{2} )} y + \frac{1}{x ( 1 + x ^{2} )}

\int_{- 1}^{1} x^{2} - x - 1 d x

(\frac{4}{3})^{^{'}}

\frac{d}{d x} (ln (5 x^{2} + 3 y^{2}))

What is the derivative of f(x)=(x^3-6x^2+3)/(x^2) ?
The derivative of f(x)=(x^3-6x^2+3)/(x^2) is (x^3-6)/(x^3)
What is the first derivative of f(x)=(x^3-6x^2+3)/(x^2) ?
The first derivative of f(x)=(x^3-6x^2+3)/(x^2) is (x^3-6)/(x^3)